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Centre de Recherches Mathématiques
CRM Proceedings and Lecture Notes
Volume ??, 2004
On the p-Adic Series
P∞
n=1
nk · n!
M. Ram Murty and Sarah Sumner
1. Introduction
Let Qp be the field of p-adic numbers. It is well known that the sum
∞
X
an ,
n=1
with an ∈ Qp converges if and only if |an |p → 0 as n → ∞ (see for example,
P∞
[6, p. 87]). Since |n!|p → 0, we see that n=1 n! converges in Qp , as well as the
P∞
related sums n=1 nk · n! for k a non-negative integer. Our goal in this paper is to
investigate these p-adic sums.
Our first question is whether
∞
X
n!
α=
n=1
is a p-adicirrational or not, a question first asked by Schikhof [9, p. 17]. We conjecture that it is. Observe however that
∞
X
n · n! = −1.
n=1
Pm
A simple proof by induction shows that 1·1!+2·2!+3·3!+· · ·+m·m! = n=0 n·n! =
(m+1)!−1. Since limm→∞ |(m + 1)!|p = 0, the desired result follows. On the other
hand, we show below that
∞
X
αk =
nk · n! = vk − uk α,
n=1
where vk , uk ∈ Z. We will prove using ideas from combinatorics and number theory
that uk 6= 0 for k ≡ 0 or 2 (mod 3). Thus, the irrationality of αk hinges on
the irrationality of α in these cases. This strengthens the observation made by
Dragovich [5] that if α is irrational, then so is αk . We suspect uk 6= 0 for every
k ≥ 2. In this paper, we will assume that α is irrational and study the sequences
uk and vk .
2000 Mathematics Subject Classification. Primary: ....; Secondary: ....
This is the final form of the paper.
c
2004
American Mathematical Society
1
2
M. RAM MURTY AND SARAH SUMNER
The sequence of integers uk defined below is of independent combinatorial interest. We hope to address the question of the non-vanishing of uk for k ≥ 2 in a
future paper. Below, we establish some relationships with Stirling numbers of the
second kind that allows us to prove non-vanishing results in some cases. It should
be clear from the context that in the summations being considered, which ones are
p-adic sums and which are summations in the field of real numbers.
2. Preliminaries
We must first recall some notation and elementary lemmas from combinatorics.
The reader who wishes a more detailed review of this material is advised to consult
Comtet [3].
The rising factorial is
hxin = x(x + 1)(x + 2) · · · (x + n − 1).
The Stirling number of the first kind, denoted by s(k, j), is defined by the rule
that (−1)k+j s(k, j) is the number of permutations of {1, . . . , k} with j cycles. The
Stirling number of the second kind, denoted by S(k, j), is the number of partitions
of {1, . . . , k} into j non-empty parts. Stirling numbers of the first kind are related
to the rising factorial by the following relation [3, p. 213].
Lemma 1.
hxij =
k
X
|s(k, j)|xj .
j=1
We will also have need for a very useful inversion lemma [3, p. 144].
Lemma 2. Let {fk } and {gk } be sequences of real numbers. Then,
fk =
k
X
S(k, j)gj ⇔ gk =
k
X
s(k, j)fj .
j=1
j=1
We recall the following basic fact which will be used below:
! ∞
!
∞
∞
X
X bn T n
X
an T n
cn T n
=
n!
n!
n!
n=0
n=0
n=0
where
cn =
n X
n
k=0
k
ak bn−k .
Bell numbers will appear below, and will be denoted Bk . Since Bk+1 is just
the total number of all partitions of the set {1, 2, . . . , k + 1} [3, p. 210], we deduce:
Lemma 3.
Bk+1 =
k+1
X
j=1
S(k + 1, j).
ON THE p-ADIC SERIES
P∞
n=1
nk · n!
3
3. A generating function for uk
P∞
P∞
We must first demonstrate that n=1 nk · n! = vk − uk α,
where α = n=1 n!
P∞
and vk and uk are integers. Notice that n=1 (n + k)! − n! is a telescoping sum
Pk
that equals − n=1 n!. For k = 2 this becomes,
∞
X
∞
X
(n + 2)! − n! =
n! (n + 2)(n + 1) − 1 ,
n=1
n=1
∞
X
∞
X
n2 · n! = −
n=1
An inductive argument now shows that
where vk , uk ∈ Z.
Observe now that for the same sum,
∞
X
n!.
n=1
P∞
n=1
nk · n! can be written as vk − uk α
∞
X
(n + k)! − n! =
(n − 1)!{(n + k)(n + k − 1) · · · n − n}
n=1
=
n=1
∞
X
∞
X
(n − 1)!hnik+1 −
n=1
n!.
n=1
Hence applying Lemma 1,
−
k
X
n! + α =
n=1
∞
X
(n − 1)!hnik+1
n=1
=
∞
X
(n − 1)!
n=1
=
k+1
X
Letting
n=1
|s(k + 1, j)|nj
j=1
|s(k + 1, j)|
j=1
P∞
k+1
X
∞
X
nj−1 n!.
n=1
nj−1 n! = vj−1 − uj−1 α gives
−
k
X
n! + α =
n=1
k+1
X
|s(k + 1, j)|{vj−1 − uj−1 α}.
j=1
If we now suppose that α is irrational, then this yields
(3.1)
k+1
X
j=1
(3.2)
k+1
X
|s(k + 1, j)|vj−1 = −
k
X
n!,
n=1
|s(k + 1, j)|uj−1 = −1.
j=1
Removing absolute values from equation (3.2) we get
(3.3)
k+1
X
j=1
(−1)j s(k + 1, j)uj−1 = (−1)k .
4
M. RAM MURTY AND SARAH SUMNER
Applying Lemma 2 to equation (3.3) yields one generating function. Let fk+1 =
(−1)k+1 uk and gj = (−1)j−1 to get
fk+1 = (−1)k+1 uk =
k+1
X
S(k + 1, j)gj
j=1
=
k+1
X
(−1)j−1 S(k + 1, j).
j=1
Thus,
Lemma 4.
(−1)k uk =
k+1
X
(−1)j S(k + 1, j).
j=1
This is a simplification of the method of solving k + 1 linear equations to
determine each uk outlined by Dragovich [5, p. 101]. Using this relation, the first
few terms in the sequence {uk } can be easily calculated:
{0, 1, −1, −2, 9, −9, −50, 267, −413, −2180, 17731, −50533, −110176,
1966797, −9938669, 8638718, 278475061, −2540956509, 9816860358 . . . }
This is the negative of sequence A014182 of Sloane [10]. Lemma 4 implies the
crucial result
Lemma 5.
uk ≡
k+1
X
S(k + 1, j) (mod 2).
j=1
Other generating functions for the uk also exist. Now we caution the reader that
below, the series being considered are not p-adic series but usual series involving
real numbers.
Lemma 6.
(−1)k uk = e
∞
X
nk+1
(−1)n .
n!
n=0
Proof. Recall the known identity [3, p. 204]
j
1 X
j
S(k + 1, j) =
(−1)r
(j − r)k+1 .
j! r=0
r
Now,
(−1)k uk =
X
(−1)j S(k + 1, j)
j≥1
=
X (−1)j X
j≥1
j!
0≤r≤j
(−1)r
j
(j − r)k+1
r
X (−1)r X (−1)j
j!
=
(j − r)k+1
r!
j! (j − r)!
r≥0
j≥r
ON THE p-ADIC SERIES
=
n=1
nk · n!
5
X 1 X (−1)j−r
(j − r)k+1
r!
(j − r)!
r≥0
=
P∞
j≥r
∞
X
X 1
nk+1
(−1)n
r! n=0 n!
r≥0
∞
X
(−1)k uk = e
nk+1
(−1)n . n!
n=0
Lemma 7.
∞
X
(−1)k uk T k
k!
k=0
= −e−e
T
+T +1
Proof.
∞
X
(−1)k uk T k
k=0
k!
=e
=e
=e
∞
∞
X
T k X (−1)n nk+1
k! n=0
n!
k=0
∞ X
∞
X
n=0 k=0
∞
X
T k nk (−1)n n
k!
n!
(−1)n T n
ne
n!
n=1
= −e
X (−1)n−1
eT (n−1) eT
(n
−
1)!
n=1
T
= −e · e−e · eT
∞
X
(−1)k uk T k
k=0
k!
= −e−e
T
+T +1
.
P
T
Since ee −1 = Bk T k /k! [3, p. 211], applying the remark made earlier about
the multiplication of exponential generating functions and Lemma 7, we see that
! ∞
!
∞
∞
X
X (−1)k uk T k
X
Bk T k
Tk
=−
k!
k!
k!
k=0
k=0
k=0
which proves
Lemma 8.
n X
n
k=0
k
Bk (−1)n−k un−k = −1.
4. Congruences for {uk }
There are several interesting congruence relations for {uk }.
Lemma 9. For every prime p, up−1 ≡ −2 (mod p).
6
M. RAM MURTY AND SARAH SUMNER
Proof. Recall that S(p, 1) = S(p, p) = 1 and that for 2 ≤ k ≤ p − 1,
S(p, k) ≡ 0 (mod p), [3, p. 219]. Observe,
p
X
(−1)p−1 up−1 =
(−1)j S(p, j),
j=1
p−1
(−1)
up−1 ≡ (−1)S(p, 1) + (−1)S(p, p)
up−1 ≡ −2
(mod p),
(mod p). Lemma 10. For every prime p, up ≡ −1 (mod p).
Proof. Substitute the recurrence S(k+1, j) = S(k, j −1)+jS(k, j), [3, p. 208]
into
(−1)p up =
p+1
X
(−1)j S(p + 1, j),
j=1
−up =
p+1
X
(−1)j S(p, j − 1) +
j=1
p+1
X
j=1
−up ≡ −up−1 + (−1)S(p, 1)
up ≡ up−1 + 1
up ≡ −2 + 1
up ≡ −1
(−1)j jS(p, j),
(mod p),
(mod p),
(mod p),
(mod p). Using the fact that for every prime p,
r
p
≡ 0 (mod p)
k
whenever 1 ≤ k ≤ pr − 1 along with Lemma 8, we conclude
Lemma 11. Let p be a prime number. For every r ≥ 1,
r
Bpr − 1 ≡ (−1)p upr
(mod p).
5. A general theorem
Theorem 1. If k ≡ 0 or 2 (mod 3) then uk 6= 0, and hence in these cases αk
will be irrational provided α is.
Since we have already shown in Lemma 5 that
uk ≡
k+1
X
S(k + 1, j) ≡ Bk+1
(mod 2),
j=1
Theorem 1 will follow from the following lemma:
Lemma 12. If k ≡ 2 (mod 3) then Bk is even, otherwise Bk is odd.
Proof. We will proceed by induction. B0 = 1, B1 = 1 and B2 = 2, so the
base cases are clear. Suppose the lemma is true for all j ≤ k. Recall the recursion
for Bk+1 ,
k X
k
Bk+1 =
Bj .
j
j=0
ON THE p-ADIC SERIES
P∞
n=1
nk · n!
7
By the induction hypothesis, if j ≡ 2 (mod 3) then Bj is even, and otherwise Bj
is odd. Hence the recursive formula becomes
X
X
X
k
k
k
≡
+
(mod 2).
j
j
j
j6≡2
(mod 3)
j≡0
(mod 3)
j≡1
(mod 3)
Let ζ = e2πi/3 be a cube root of unity. From the binomial theorem, we see
k X
k
j
j=0
k X
j=0
k j j
ζ x = (1 + ζx)k ,
j
k X
k
j
j=0
xj = (1 + x)k ,
ζ 2j xj = (1 + ζ 2 x)k .
Adding these together we get
k X
k
j=0
j
xj (1 + ζ j + ζ 2j ) = (1 + x)k + (1 + ζx)k + (1 + ζ 2 x)k .
Let x = 1.
k X
k
j=0
j
(1 + ζ j + ζ 2j ) = 2k + (1 + ζ)k + (1 + ζ 2 )k .
Recall
j
1+ζ +ζ
2j
(
3 if j ≡ 0 (mod 3)
=
0 otherwise.
So
X
3
j≡0
(mod 3)
k
= 2k + (1 + ζ)k + (1 + ζ 2 )k
j
= 2k + (1 + ζ)k + (1 + ζ −1 )k
= 2k + 2<(1 + ζ)k
= 2k − 2<(ζ 2k ).
Let us note that
(5.1)
ζ 2k


1
= ζ

 2
ζ
if k ≡ 0 (mod 3)
if k ≡ 2 (mod 3)
if k ≡ 1 (mod 3).
8
M. RAM MURTY AND SARAH SUMNER
Now for the other sum. Consider
k X
k
j
j=0
k X
j=0
k j−1 j
ζ
x = ζ −1 (1 + ζx)k ,
j
k X
k
j
j=0
xj = (1 + x)k ,
ζ 2j−2 xj = ζ −2 (1 + ζ 2 x)k .
Adding these together gives
k X
k j
x (1 + ζ j−1 + ζ 2j−2 ) = (1 + x)k + ζ −1 (1 + ζx)k + ζ −2 (1 + ζ 2 x)k .
j
j=0
Let x = 1
k X
k
j=0
j
(1 + ζ j−1 + ζ 2j−2 ) = 2k + ζ −1 (1 + ζ)k + ζ −2 (1 + ζ 2 )k .
Now
1+ζ
j−1
+ζ
2j−2
(
3
=
0
if j − 1 ≡ 0
otherwise.
(mod 3)
.
Hence,
X
3
j≡1
(mod 3)
k
= 2k + ζ −1 (1 + ζ)k + ζ(1 + ζ −1 )k
j
= 2k + 2<ζ(1 + ζ −1 )k
= 2k + 2<ζ(1 + ζ 2 )k
= 2k + 2<ζ(−ζ)k
= 2k + (−1)k 2<(ζ k+1 ).
Now
(5.2)
ζ
k+1


1
= ζ

 2
ζ
if k + 1 ≡ 0 (mod 3)
if k + 1 ≡ 1 (mod 3)
if k + 1 ≡ 2 (mod 3).
Combining the information in equations (5.1) and (5.2), we see that if k ≡ 1
(mod 3) then Bk+1 ≡ 0 (mod 2), and if k 6≡ 1 (mod 3) then Bk+1 ≡ 1 (mod 2)
proving Theorem 1.
6. Concluding remarks
The non-vanishing of uk is a conjecture of Wilf (see [7]). In [11], it is proved
that the number of k ≤ x with uk = 0 is O(x2/3 ).
Other questions deserving attention linked to the study of {uk } concern the
existence and properties of p-adic interpolation of functions of the form
X
f (s) =
ns · n!.
ON THE p-ADIC SERIES
P∞
n=1
nk · n!
9
If the sum is taken over all odd numbers n, then it can be shown that f (s) has
a 2-adic interpolation, which raises the question of whether {uk } and {vk } have
2-adic limits or not. Similar questions can be raised about p-adic interpolation and
limits for odd primes, which seem to be more involved cases.
References
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theory, Academic Press, Toronto, 1966.
P
2. E. B. Burger and T. Struppeck, Does ∞
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(1993), no. 3, 95–105.
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, On some p-adic series with factorials, p-adic Functional Analysis, (W. H. Schikhof,
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Dekker, Inc., New York, 1997, pp. 95-105.
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527–532.
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Press, Cambridge, 1984.
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50 (2003), no. 8, 912–915; http://www.research.att.com/∼njas/sequences.
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Department of Mathematics and Statistics, Queen’s University, Kingston, ON K7L
3N6, Canada.
E-mail address: [email protected]
General Dynamics Canada Ltd., Ottawa, ON K2H 5B7, Canada.
E-mail address: [email protected]